Answer on Question #57287 Physics / Molecular Physics | Thermodynamics
Two cylinders of gas are identical in every way, except one contains oxygen gas, the other contains helium gas. Both hold the same volume of gas at STP and are closed by a movable piston at one end. Both gases are now compressed adiabatically to one-third their original volume. Which gas will show the greater temperature increase, or will they both show the same temperature increase?
Given:
i1=5 - degree of freedom of two atomic oxygen (O2)
i2=3 - degree of freedom of one atomic helium (He)
V0,P0,T0 - initial state of each gas
V1=V2=3V0 - volumes after compression
Adiabatically (Q=0)
ΔT1ΔT2−?Solution:
Adiabatic process can be described by the formulas:
PVγ=const, or TVγ−1=const, where γ=ii+2
It's convenient to use the second one in our problem. We apply this formula to the states before and after compression of each gas respectively.
T0V0γ1−1=T1V1γ1−1T0V0γ2−1=T2V2γ2−1γ1=57;γ2=35
From where
T1=T0(V1V0)γ1−1=T03γ1−1ΔT1=T0(3γ1−1−1)ΔT2=T0(3γ2−1−1)ΔT1ΔT2=3γ1−1−13γ2−1−1=32/5−132/3−1≈1.96Answer:
ΔT2=1.96ΔT1. Increase in temperature of helium is 1.96 times larger than that of oxygen gas.
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